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Question: Solve \[{{\cot }^{-1}}(3)+\cos e{{c}^{-1}}(\sqrt{5})=\] A.\[\dfrac{\pi }{3}\] B.\[\dfrac{\pi }{4...

Solve cot⁡−1(3)+cos⁡ec−1(5)={{\cot }^{-1}}(3)+\cos e{{c}^{-1}}(\sqrt{5})=
A.π3\dfrac{\pi }{3}
B.π4\dfrac{\pi }{4}
C.π6\dfrac{\pi }{6}
D.π2\dfrac{\pi }{2}

Explanation

Solution

Hint : An equation which involves trigonometric ratio of any angle is said to be a trigonometric identity if it is satisfied for all values for which the given trigonometric ratios are defined. Trigonometric functions are periodic functions and all trigonometric functions are not bijections. Consequently their inverse does not exist. If no branch of an inverse trigonometric function is given, then it means that the principal value branch of the function. An inverse function reverses the direction of the original function.

Complete step-by-step answer :
Let us assume that x=cos⁡ec−1(5)x=\cos e{{c}^{-1}}(\sqrt{5})
By solving we get
⇒cos⁡ecx=5\Rightarrow \cos ecx=\sqrt{5}
Using the trigonometric identity cos⁡ec2θ=1+cot⁡2θ\cos e{{c}^{2}}\theta =1+{{\cot }^{2}}\theta
Rearranging the identity we get
cot⁡2x=cos⁡ec2x−1{{\cot }^{2}}x=\cos e{{c}^{2}}x-1
Substituting the value we get
cot⁡2x=(5)2−1{{\cot }^{2}}x={{(\sqrt{5})}^{2}}-1
Further simplifying we get
cot⁡2x=5−1{{\cot }^{2}}x=5-1
On solving we get
cot⁡2x=4{{\cot }^{2}}x=4
Taking square root on both sides we get
cot⁡x=2\cot x=2
Further rearranging we get
x=cot⁡−1(2)x={{\cot }^{-1}}(2)
Equating the values we get
cos⁡ec−1(5)=cot⁡−1(2)\cos e{{c}^{-1}}(\sqrt{5})={{\cot }^{-1}}(2)
So we can write the equation as
cot⁡−1(3)+cos⁡ec−1(5)=cot⁡−1(3)+cot⁡−1(2){{\cot }^{-1}}(3)+\cos e{{c}^{-1}}(\sqrt{5})={{\cot }^{-1}}(3)+{{\cot }^{-1}}(2)
The above equation can be written as
cot⁡−1(3)+cos⁡ec−1(5)=tan⁡−113+tan⁡−112{{\cot }^{-1}}(3)+\cos e{{c}^{-1}}(\sqrt{5})={{\tan }^{-1}}\dfrac{1}{3}+{{\tan }^{-1}}\dfrac{1}{2}
Now we will use the trigonometric formula given as
tan⁡−1A+tan⁡−1B=tan⁡−1(A+B1−AB){{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right)
Using this trigonometric formula we get
cot⁡−1(3)+cos⁡ec−1(5)=tan⁡−1(13+121−13×12){{\cot }^{-1}}(3)+\cos e{{c}^{-1}}(\sqrt{5})={{\tan }^{-1}}\left( \dfrac{\dfrac{1}{3}+\dfrac{1}{2}}{1-\dfrac{1}{3}\times \dfrac{1}{2}} \right)
Further simplifying we get
cot⁡−1(3)+cos⁡ec−1(5)=tan⁡−1(55){{\cot }^{-1}}(3)+\cos e{{c}^{-1}}(\sqrt{5})={{\tan }^{-1}}\left( \dfrac{5}{5} \right)
On solving we get
cot⁡−1(3)+cos⁡ec−1(5)=tan⁡−1(1){{\cot }^{-1}}(3)+\cos e{{c}^{-1}}(\sqrt{5})={{\tan }^{-1}}(1)
Using trigonometric ratios and angles concept to solve further we get
cot⁡−1(3)+cos⁡ec−1(5)=tan⁡−1(tan⁡π4){{\cot }^{-1}}(3)+\cos e{{c}^{-1}}(\sqrt{5})={{\tan }^{-1}}(\tan \dfrac{\pi }{4})
As we know that
tan⁡−1(tan⁡θ)=θ{{\tan }^{-1}}(\tan \theta )=\theta
Applying this we get
cot⁡−1(3)+cos⁡ec−1(5)=π4{{\cot }^{-1}}(3)+\cos e{{c}^{-1}}(\sqrt{5})=\dfrac{\pi }{4}
Therefore, option BB is the correct answer.
So, the correct answer is “Option B”.

Note : Before solving the trigonometric problems, one must be familiar with the trigonometric ratios, trigonometric identities, inverse trigonometric functions and trigonometric applications. The word ‘trigonometry’ is derived from the Greek words ‘tri’ which means three, ‘gon’ (means sides) and ‘metron’ (means measure). Some ratios of the sides with respect to its acute angles, called trigonometric ratios of the angle.